This two-dimensional shape can be folded onto the surface of a regular polyhedron (one of the five platonic solids) in a way that perfectly covers the entire surface of the polyhedron with no gaps and no overlaps.

How can this be done, and which one of the platonic solids can it be done to?

1 Answer
1

The triangular symmetry shows it has to be a polyhedron made of triangles. The icosahedron seems too complex compared to the fractal shown. It must be the tetrahedron or the octahedron.

After some playing around with Acorn, I came up with the answer:

It is the octahedron.

You see in red the ouline of the unfolded octahedron. The black regions A, B and C nicely fill in the white regions with the same name. The three black regions at the tips combine to form the face opposite to the center.
That face is actually missing from the red outline.

$\begingroup$The final face consists of a 1/4 of a triangle at each tip, with a 1/4 of that triangle on an edge, with a 1/4 of that triangle on an edge, etc.... (1/4+1/16+...=1/(1-1/4)-1=1/3 and we have 3 such constructs).$\endgroup$
– JMPJun 29 at 10:56

$\begingroup$@JonMarkPerry The fractal you link to has a lot of interesting similarities to mine, but note that it's not the same fractal. It can be generated using the same subdivision rules as mine, but it has a different starting point.$\endgroup$
– plasticinsectJun 29 at 16:52

2

$\begingroup$If you go to the tessellation image with the animation, your fractal is 4 fudgeflakes sandwiched together (the black one is centre to the red, orange and cyan ones).$\endgroup$
– JMPJun 29 at 16:54

$\begingroup$@JonMarkPerry Aha! I see that now that you point it out. Very interesting. This makes perfect sense though, because the starting point of my fractal is made up of four copies of the starting point of the fudgeflake fractal (for the third construction method given on that page).$\endgroup$
– plasticinsectJun 29 at 17:02